## Question 4:

### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 - x + 2$ has discriminant $-7$, so $x^2 - x + 2 \neq 0$ and when e.g. $x=0$, $x^2 - x + 2 > 0$ so positive for all $x$ | E2,1,0 | Discriminant $< 0$ shown **and** sign of $x^2 - x + 2$ or curve position discussed. Allow complex roots found, with discussion |
| $\mathbf{OR}\ x^2 - x + 2 = (x - \tfrac{1}{2})^2 + \tfrac{7}{4} \geq \tfrac{7}{4} > 0$ for all $x$ | E2,1,0 | Completing the square and minimum value discussed |
| $\mathbf{OR}$ using $y = x^2 - x + 2$: $\frac{dy}{dx} = 2x-1 = 0$ when $x = \frac{1}{2}$ and $y = \frac{7}{4}$; $\frac{d^2y}{dx^2} = 2 > 0$. Hence $y \geq \frac{7}{4} > 0$ for all $x$ | E2,1,0 | Calculus, showing minimum value $> 0$ |
| | **[2]** | |

### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{2x}{x^2 - x + 2} > x$ | | |
| $\Rightarrow 2x > x^3 - x^2 + 2x$ | M1 | Valid attempt to eliminate fraction. Or combine to one fraction $>$ or $<0$ |
| $\Rightarrow 0 > x^3 - x^2 \Rightarrow 0 > x^2(x-1)$ | M1 | Simplification and factors |
| $0, 1$ critical values | A1 | Both, no other values given |
| $x < 1$ | A1 | |
| $\Rightarrow x < 0$ or $0 < x < 1$ or $x < 1, x \neq 0$ | A1 | cao |
| | **[5]** | |
| $\mathbf{OR}$ Graphical approach by sketching $y = \frac{2x}{x^2-x+2}$ and $y=x$ or $y = \frac{2x}{x^2-x+2} - x$ | M2,1,0 | Accuracy of sketch |
| Critical values $0$ and $1$ | A1 | Both |
| $x < 1$ | A1 | |
| $\Rightarrow x<0$ or $0<x<1$ or $x<1, x\neq 0$ | A1 | cao |
| | **[5]** | |

---